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burhan101

  • 3 years ago

Trig identity.. I don't understand this one at all

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  1. burhan101
    • 3 years ago
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    how do i make the left side into one big fraction

  2. burhan101
    • 3 years ago
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    \[\huge \frac{ 1 }{ sinx+1 } - \frac{ 1 }{ sinx-1 } = \frac{ 2 }{ 1-\sin^2x}\]

  3. burhan101
    • 3 years ago
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    typed it wrong wooops

  4. dpaInc
    • 3 years ago
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    get common denominators on the left side: \(\huge \frac{ 1 }{ sinx+1 }\cdot (\color {red}{\frac{ sinx-1 }{ sinx-1 }}) - \frac{ 1 }{ sinx-1 }\cdot( \color {red}{\frac{ sinx+1 }{ sinx+1 }}) = \frac{ 2 }{ 1-\sin^2x} \)

  5. dpaInc
    • 3 years ago
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    so the left side in simplified form will be: \(\huge \frac{(sinx-1)-(sinx+1)}{sin^2x-1} \) can you simplify this?

  6. burhan101
    • 3 years ago
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    no that's where i get stuck how would i subtract the numerator

  7. dpaInc
    • 3 years ago
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    just use distributive property to get rid of the parenthesis...

  8. burhan101
    • 3 years ago
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    @dpaInc i get this \[\huge \frac{ \sin^2x+sinx-sinx-1 }{ sinx^2-1 }\]

  9. dpaInc
    • 3 years ago
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    how did you get that? all you're doing is subtracting....: \(\huge \frac{(sinx-1)-(sinx+1)}{sin^2x-1} \) =\(\huge \frac{1\cdot (sinx-1)-1\cdot (sinx+1)}{sin^2x-1} \) =\(\huge \frac{sinx-1-sinx-1}{sin^2x-1} \) =\(\huge \frac{-2}{sin^2x-1} \) =\(\huge \frac{2}{1-sin^2x} \)

  10. burhan101
    • 3 years ago
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    sinx-1-sinx-1 = -2 is that a trig rule or what ? :S

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